Help 87% Mon 1 CHEM1100.04-Section 4 -Practice Exam3 Topics | 21 of 36 17 18 19

Question

Help 87% Mon 1 CHEM1100.04-Section 4 -Practice Exam3 Topics | 21 of 36 17 18 19 20 22 2324 25 2627282930 31 34 35 that A chemistry student needs to standardize a fresh solution of sodum hydroxide. She carefuily weighs out 102, mg of oxalic acid (H,C204), a diprotic acid can be purchased inexpensively in high purity, and dissolves it in hydraxide solution. When the titration reaches 250, mL of distilled water. The student then titrates the oxalic acid solution with her sodium s the equivalence point, the student finds she has used 133.9 mL of sodium hydroxide solution Calculate the molarity of the student's sodium hydroxide solution. Be sure your answer has the correct number of significant digits.

Explanation / Answer

The given mass of oxalic acid = 108 mg = 0.108 g

The molar mass of oxalic acid = 100 g mol-1

i.e. The no. of moles of oxalic acid = 0.108 g/100 g mol-1 = 0.00108 mol

The concentration of oxalic acid = 0.00108 mol / (250*10-3 L) = 0.00432 M

At equivalence point, the no. of moles of oxalic acid = no. of moles of NaOH

i.e. The no. of moles of NaOH = 0.00108 mol

And the volume of NaOH used = 133.9 mL

Therefore, the molarity of NaOH = 0.00108 mol / 133.9 mL = 8.1*10-6 M

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